Abstract

The thermodynamic equilibrium of solid–liquid and solutions of the mixed system of sodium and hydrogen fluorides were investigated at various temperatures. The chemical speciation of the NaF-HF in aqueous solutions was described; which the considered equilibrium is dissociating into the species Na+(aq), H+(aq), HF(aq), F-(aq), HF2-(aq) and H2F3-(aq). The species molalities are strongly dependent on the concentration in the solution, therefore directly influence the thermodynamic properties. Neglecting this concentration dependence leads to erroneous osmotic coefficients for weak electrolyte solutions. Then, the water activity measurements of the mixed system yHF+1-yNaFaq were performed at various temperatures from 298.15 K to 353.15 K using the hygrometric method. These measurements were made as a function of total molality ranging from dilution to saturation for different ionic-strength fractions y with respect to the support solute HF with y = 0.20, 0.50, and 0.80. The mixed aqueous electrolytes were considered as a polyelectrolyte system [Na+-HF0-H+-F--HF2--H2F3--H2O]. Based on the ion interaction model, the thermodynamic model was developed to describe the behavior of sodium and hydrogen fluorides in aqueous solutions at different temperatures. The species molalities of the HF(aq), Na+(aq), H+(aq), F-(aq), HF2-(aq) and H2F3-(aq) of the system were evaluated. The new ion interaction parameters ΦH+, Na+ and ΨH+, Na+,F- were determined at various temperatures and used to calculate the species activity coefficients as a function of total molality for different ionic-strength fractions (y). The developed model exhibits a high consistency for the treatment of polyelectrolyte system. The results allow describing the behavior of the mixed NaF-HF in aqueous solutions. The solubility of the sodium fluoride in hydrofluoric acid aqueous solutions were also performed, and the formed solid phases were characterized by X-ray diffraction analysis. The obtained crystal forms were identified to NaF (s) and NaHF2(s) for y = 0.20 and y= (0.50 and 0.80) respectively.

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